POLLUTION CONTROLSEDIMENTATION

Sedimentation is the tendency for particles in suspension or molecules in solution to settle out of the fluid in which they are

entrained, and come to rest against a wall. This is due to their motion through the fluid in response to the forces acting on them:

these forces can be due to gravity, centrifugal acceleration or

electromagnetism

Sedimentation

Sedimentation basin

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Usually rectangular or circular with either a radial or upward water flow pattern

Can be divided into 4 zones: inlet, settling, outlet and sludge storage

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INLET ZONE

OUTLET ZONESLUDGE

ZONE

SETTLING ZONE

Horizontal flow clarifier

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SLUDGE ZONE

INLET ZONE

SETTLING ZONE

OUTLET ZONE

Inlet zone

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Purpose : to evenly distribute the flow and suspended particles across the cross section of the settling zone

Consists of a series of inlet pipes and baffles placed about 1m into the tank and extending the full depth of the tank

End of inlet zone: when the flow pattern is evenly distributed and the water velocity slowed to the design velocity of the sedimentation zone

Settling zone

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In an accurate design, the inlet and settling zones are each designed separately and their lengths added together

After passing through the inlet zone, water enters the settling zone where water velocity is greatly reduced.

This is where the bulk of settling occurs and this zone will make up the largest volume of the sedimentation basin.

For optimal performance, the settling zone requires a slow, even flow of water. The settling zone may be simply a large area of open water.

Outlet zone

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To remove the settled water from the basin without carrying away any of the floc particles

controls the amount of water flowing out of the sedimentation basin

A fundamental property of water is that the velocity of flowing water is proportional to the flow rate divided by the area through which the water flows, that is:

υ =

where υ = water velocity, m/s Q = water flow, m3/s Ac = cross-sectional area, m2

Sludge storage zone

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Depends upon the method of cleaning, the frequency of cleaning and the quantity of sludge estimated to be produced

The sludge zone is found across the bottom of the sedimentation basin where the sludge is collected temporarily . Velocity in this zone should be very slow to prevent resuspension of sludge.

A drain at the bottom of the basin allows the sludge to be easily removed from the tank. The tank bottom should slope toward the drains to further facilitate sludge removal. In some plants, sludge removal is achieved continuously using automated equipment. In other plants, sludge must be removed manually.

Scouring

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Within the sedimentation tank, the flow is going through a very large area, consequently the velocity is slow

To remove the water from the basin quickly, it is desirable to direct the water into a pipe or small channel for easy transport, which significantly produce higher velocity

Scouring which is a phenomenon of washing out the floc will happen if a pipe were to be placed at the end of the sedimentation basin (ꜛ velocity)

It is desirable to first put a series of troughs, called weirs, which provide a large area for the water to flow through and minimize the velocity near the outlet zone

Weir

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Weir

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The length of weir required is a function of the type of solids

The heavier the solids, the harder it is to scour them and the higher the allowable outlet velocity

Heavier particles require a shorter length of weir than do light particles

Sedimentation concept

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2 important term: the particle (floc) settling velocity, vs

Overflow rate, vo

If vs > vo, 100% removal of particle

If vs < vo, 0% removal

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Overflow rate: the water is flowing over the top of the tank into the weir system.

vo = volume/ time = (depth)(surface area) = depth

surface area (time)(surface area) time

vo = V/θ = (h)(As) = h

As (θ)(As) θ

The particle removal is independent of the depth of the sedimentation tank

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3 assumptions of an ideal horizontal sedimentation tank:

1. Particle and velocity vectors are evenly distributed across the tank cross section. This is the function of the inlet zone

2. The liquid moves an ideal slug down the length of the tank

3. Any particle hitting the bottom of the tank is removed

Horizontal sedimentation tank

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Some particle with vs less than vo will removed

Percentage of particle removed, PP = 100 vs

vo

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Example

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Remove particle that have vs 0.1 mm/s, 0.2 mm/s and 1 mm/s with vo =17 m3/d.m2 0.1 mm/s

17 m3/d.m2 = 17 m/d (1000 mm/m) = 0.2mm/s (86400 s/d)

Vs < Vo . So, some particle will be removed. 0.2 mm/s

Vs = Vo . Ideally will be 100% removed.

1 mm/s Vs > Vo. 100% of the particle easily removed.

Determination of vs

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1. Ideally: vs > vo

2. Determine vs , then set vo to be smaller than vs

3. Vs varies with type of particle, classified according to settling properties.

Type 1 Sedimentation

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Particle settle discretely at constant settling velocity.

Settle as individual, do not flocculate or stick to other particles.

Example: sand and grit material. Application: 1) removal of sand prior to coagulation 2) In grit chamber 3) Settling of sand particle during cleaning of rapid sand filter

Type 2 Sedimentation

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Particle that flocculate during sedimentation.

Constant change in size and settling velocity.

Generally vs increasing.Example: Alum and ironApplication: 1) Primary sedimentation 2) Settling tank in trickling filtration

Type 3 Sedimentation

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Particles are at high concentration (> 1000 mg/L)

Settle as mass with distinct clear zone and sludge zone.

Application: 1) Lime softening sedimentation 2) Activated sludge sedimentation 3) Sludge thickeners

Determination of vo for type 1 sedimentation

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Particle

FD FB

FG

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FG = gravitational force

FB = buoyancy force

FD = drag force

FG = gravitational force

ρS = density of particle, kg/m3

ρ = density of fluid, kg/m3

g = accelaration due to gravity, m/s2

Vp = volume of particle, m3

CD = drag coefficient

Ap = cross sectional area of particle, m2

v = velocity of particle, m/s

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Driving force for accelaration of particle

Drag force is equal to the driving force, the particle velocity reaches constant value called terminal settling velocity

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For spherical particles with a diameter = d,

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For spherical particles with a diameter = d,

Determining CD

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CD depend on flow regime surrounding particle:

Laminar: fluid moves in layer or gliding smoothly above adjacent layer with only molecular interchange of momentum

Turbulent: fluid motion is erratic with violent transverse interchange of momentum.

Flow is describe using dimensionless ratio, Reynold No, R

Determining R

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For sphere moving through a liquid:

R= Reynold number

d = diameter of sphere, m

vS = density of sphere, kg/m3

v = kinematic viscosity, m2/s = μ/ρ

ρ = density of fluid, kg/m3

μ = dynamic viscosity, Pa.s (appendix A)

Relation of CD with R

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No Reynold Number CD

1 R > 104 Around 0.4

2 R < 0.5 CD = 24/R

3 0.5<R< 104 CD = 24/R+ 3/R1/2 + 0.34

Stokes’ Law

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μ = dynamic viscosity, Pa.s (appendix A)

Only applied to spherical particle falling under laminar conditions.

Derivation of Stokes Law

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Fg = (4/3)r3ρpg                     (1a)FA = (4/3)r3ρfg,                    (1b)

Fr = Fg ─ FA = (4/3)r3g(ρ p  ─ ρf.)                       (2)

As soon as the sphere starts moving there is a third force, the frictional force Ff of the fluid. Its direction is opposite to the direction of motion. The total resulting force is:

   Ftot  = Fr, ─ Ff.                      (3)

As long as Ftot  is positive, the velocity increases. However, Ff  is dependent on the velocity. Over a large range of velocities The frictional force is proportional to the velocity (v):

   Ff = 6πrμv                            (4)

where  μ  is the dynamic fluid viscosity

Derivation of Stokes Law

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After some time the velocity does not increase anymore but becomes constant. Then equilibrium is reached. In other words, Fr is canceled by Ff and so Ftot  = 0. From now on the particle has a constant velocity. The equilibrium or setting velocity vs can be calculated from (2), (3) and (4) with Ftot = Fr, ─ Ff  = 0. The result is:

  vs = (2/9)r2g(ρp ─ ρf)/ μ       (5).

μ = dynamic viscosity, Pa.s (appendix A)

This equation only holds under ideal conditions, such as a very large fluid medium, a very smooth surface of the sphere and a small radius.

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Now that we have determine vs, we can set vo

Vo is recommended to be set at 0.33 to 0.7 times depending upon efficiency desired.

Flocullant Sedimentation Lab or Pilot Data

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The Stokes equation cannot be used because the flocculating are continually changing in size and shape, and when water is entrapped in the floc, in specific gravity.

The concentration of suspended solids is determined for each sample and the percent removal is calculated:

%)100(1%o

t

C

CR

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Where R% = percent removal at one depth and

time, %

Ct = concentration at time, t, and given depth

Co = initial concentration, mg/L

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Each intersection point of an isoconcentration line and the bottom of the column defines an overflow rate (vo ):

where H = height of column, m ti = time defined by intersection of

isoconcentration line and bottom of column (x-axis) where the subscript, i, refers to the first, second, third, etc., intersection points, d

io t

H

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RTa = Ra + H1 (Rb - Ra) + H2 (Rc - Rb) + ….

H Hwhere RTa = total fraction removed for settling time,

ta

Ra, Rb, Rc = isoconcentration fractions a, b, c, etc.

H1 ,H2 = the midpoints between isoconcentration lines used to calculate the fraction of solids removed.

Note : Eckenfelder (1980) recommends that scale-up factors of 0.65 for

overflow rate and 1.75 for detention time be used to design the tank.

EXAMPLE 4-23

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The city of Urbana is planning to install a new water treatment plant. Design a settling tank to remove 65% of the influent suspended solids from their design flow of 0.5m3/s. A batch-settling test using a 2.0 m column and coagulated water from their existing plant yielded the following data :

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Percent removal as a function of time and depth

Sampling Time, min

5 10 20 40 60 90 120Depth, m

0.5 41 50 60 67 72 73 76

1.0 19 33 45 58 62 70 74

2.0 15 31 38 54 59 63 71

20

40

38

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SOLUTION

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The plot is shown in Figure 4-40.Calculate the overflow rate for each

intersection point.For example, for the 50% line,

(1440 min/d) = 82.3 m/d

Interpolation based on table

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The corresponding removal fraction is

RT50 = 50 + 1.5 (55-50) + 0.85 (60-55) +

2.0 2.0 0.60 (65-60) + 0.40 (70 - 65) +

2.0 2.0 0.20 (75 - 70) + 0.05 (100 -75)

2.0 2.0

= 59.5 or 60%

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This calculation is repeated for each isoconcentration line that intersects the x-axis except the last ones for which data are too sparse (30, 40, 50, 55, 60 and 65%)

Using those calculation, construct two graphs (Figures 4-41 and 4-42)

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49

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From those graphs (65 percent removal),bench-scale detention time = 54 min

overflow rate = 50 m/dApplying the scale-up factors yields

to = (54 min)(1.75) = 94.5 or 95 min

υo = (50 m/d)(0.65) = 32.5 m/d

Zone Sedimentation Lab Data

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The design overflow is again set at about 0.5-0.7 × the lab value.

A technique has been developed to determine settling velocities of coagulant flocs from jar test data (Hudson, 1981)

Typical detention times for waters coagulated with alum or iron salts are on the order of 2-8 hours. In lime-soda softening plants, the detention times range from 4-8 hours (Reynolds and Richards, 1996)

EXAMPLE 4-24

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Determine the surface area of a settling tank for the city of Urbana’s 0.5 m3/s design overflow rate found in Example 4-23. Compare this surface area with that which results from assuming a typical overflow rate of 20 m3/d. Find the depth of the clarifier for the overflow rate and detention time found in Example 4-23.

SOLUTION

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a) Find the surface area. First change the flow rate to compatible

units : (0.5 m3/s) (86400 s/d) = 43200 m3/d

The surface area :As = 43200 m3/d = 1329.23 or 1330 m2

32.5 m3/d.m2

Using the overflow rate from Example 4-23.

OR

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As = 43200 m3/d = 2160 m2

20 m3/d.m2

Using the conservative value given

Note: the use of conservative data would, in this case

result in 60% overdesign of the tank area

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Common length-to-width ratios for settling are between 2:1 and 5:1, and lengths seldom exceed 100m. A minimum of two tanks is always provided.

Assuming 2 tanks, each with a width of 12 m, a total surface area of 1330m2 would imply a tank length of

this meets our length-to-width ratio 5:1

m 55or 4.55 wide)m tanks)(12(2

1330Length

2

m

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b) Find the tank depth.First, find the total volume tank from Eqn 2-27 using the detention time of 95 mins from Example 4-23 :

V = (0.5 m3/s) (95 min) (60 s/min)= 2850 m3

This would be divided into 2 tanks as noted above.

Depth = 2850 m3 = 2.1428 or 2 m 1330 m2